Integrand size = 66, antiderivative size = 25 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=1+(5+x) \left (5-e^x+2 x+\log ^2\left (2 \log \left (x^2\right )\right )\right ) \] Output:
1+(5+x)*(ln(2*ln(x^2))^2+5-exp(x)+2*x)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=-e^x (5+x)+x (15+2 x)+(5+x) \log ^2\left (2 \log \left (x^2\right )\right ) \] Input:
Integrate[((15*x + 4*x^2 + E^x*(-6*x - x^2))*Log[x^2] + (20 + 4*x)*Log[2*L og[x^2]] + x*Log[x^2]*Log[2*Log[x^2]]^2)/(x*Log[x^2]),x]
Output:
-(E^x*(5 + x)) + x*(15 + 2*x) + (5 + x)*Log[2*Log[x^2]]^2
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )+(4 x+20) \log \left (2 \log \left (x^2\right )\right )+\left (4 x^2+e^x \left (-x^2-6 x\right )+15 x\right ) \log \left (x^2\right )}{x \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )+4 x^2 \log \left (x^2\right )+15 x \log \left (x^2\right )+4 x \log \left (2 \log \left (x^2\right )\right )+20 \log \left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )}-e^x (x+6)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \log ^2\left (2 \log \left (x^2\right )\right )dx+4 \int \frac {\log \left (2 \log \left (x^2\right )\right )}{\log \left (x^2\right )}dx+2 x^2+5 \log ^2\left (2 \log \left (x^2\right )\right )+15 x+e^x-e^x (x+6)\) |
Input:
Int[((15*x + 4*x^2 + E^x*(-6*x - x^2))*Log[x^2] + (20 + 4*x)*Log[2*Log[x^2 ]] + x*Log[x^2]*Log[2*Log[x^2]]^2)/(x*Log[x^2]),x]
Output:
$Aborted
Time = 0.60 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64
method | result | size |
parallelrisch | \(x {\ln \left (2 \ln \left (x^{2}\right )\right )}^{2}+2 x^{2}-{\mathrm e}^{x} x +5 {\ln \left (2 \ln \left (x^{2}\right )\right )}^{2}+15 x -5 \,{\mathrm e}^{x}\) | \(41\) |
risch | \(\left (5+x \right ) {\ln \left (4 \ln \left (x \right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}\right )}^{2}+2 x^{2}-{\mathrm e}^{x} x +15 x -5 \,{\mathrm e}^{x}\) | \(59\) |
Input:
int((x*ln(x^2)*ln(2*ln(x^2))^2+(20+4*x)*ln(2*ln(x^2))+((-x^2-6*x)*exp(x)+4 *x^2+15*x)*ln(x^2))/x/ln(x^2),x,method=_RETURNVERBOSE)
Output:
x*ln(2*ln(x^2))^2+2*x^2-exp(x)*x+5*ln(2*ln(x^2))^2+15*x-5*exp(x)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx={\left (x + 5\right )} \log \left (2 \, \log \left (x^{2}\right )\right )^{2} + 2 \, x^{2} - {\left (x + 5\right )} e^{x} + 15 \, x \] Input:
integrate((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6* x)*exp(x)+4*x^2+15*x)*log(x^2))/x/log(x^2),x, algorithm="fricas")
Output:
(x + 5)*log(2*log(x^2))^2 + 2*x^2 - (x + 5)*e^x + 15*x
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=2 x^{2} + 15 x + \left (- x - 5\right ) e^{x} + \left (x + 5\right ) \log {\left (2 \log {\left (x^{2} \right )} \right )}^{2} \] Input:
integrate((x*ln(x**2)*ln(2*ln(x**2))**2+(20+4*x)*ln(2*ln(x**2))+((-x**2-6* x)*exp(x)+4*x**2+15*x)*ln(x**2))/x/ln(x**2),x)
Output:
2*x**2 + 15*x + (-x - 5)*exp(x) + (x + 5)*log(2*log(x**2))**2
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=4 \, x \log \left (2\right )^{2} + {\left (x + 5\right )} \log \left (\log \left (x\right )\right )^{2} + 2 \, x^{2} - {\left (x - 1\right )} e^{x} + 4 \, {\left (x \log \left (2\right ) + 5 \, \log \left (2\right )\right )} \log \left (\log \left (x\right )\right ) + 15 \, x - 6 \, e^{x} \] Input:
integrate((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6* x)*exp(x)+4*x^2+15*x)*log(x^2))/x/log(x^2),x, algorithm="maxima")
Output:
4*x*log(2)^2 + (x + 5)*log(log(x))^2 + 2*x^2 - (x - 1)*e^x + 4*(x*log(2) + 5*log(2))*log(log(x)) + 15*x - 6*e^x
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=x \log \left (2 \, \log \left (x^{2}\right )\right )^{2} + 2 \, x^{2} - x e^{x} + 5 \, \log \left (2 \, \log \left (x^{2}\right )\right )^{2} + 15 \, x - 5 \, e^{x} \] Input:
integrate((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6* x)*exp(x)+4*x^2+15*x)*log(x^2))/x/log(x^2),x, algorithm="giac")
Output:
x*log(2*log(x^2))^2 + 2*x^2 - x*e^x + 5*log(2*log(x^2))^2 + 15*x - 5*e^x
Time = 3.72 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=15\,x-{\mathrm {e}}^x\,\left (x+5\right )+{\ln \left (2\,\ln \left (x^2\right )\right )}^2\,\left (x+5\right )+2\,x^2 \] Input:
int((log(x^2)*(15*x - exp(x)*(6*x + x^2) + 4*x^2) + log(2*log(x^2))*(4*x + 20) + x*log(2*log(x^2))^2*log(x^2))/(x*log(x^2)),x)
Output:
15*x - exp(x)*(x + 5) + log(2*log(x^2))^2*(x + 5) + 2*x^2
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {\left (15 x+4 x^2+e^x \left (-6 x-x^2\right )\right ) \log \left (x^2\right )+(20+4 x) \log \left (2 \log \left (x^2\right )\right )+x \log \left (x^2\right ) \log ^2\left (2 \log \left (x^2\right )\right )}{x \log \left (x^2\right )} \, dx=-e^{x} x -5 e^{x}+{\mathrm {log}\left (2 \,\mathrm {log}\left (x^{2}\right )\right )}^{2} x +5 {\mathrm {log}\left (2 \,\mathrm {log}\left (x^{2}\right )\right )}^{2}+2 x^{2}+15 x \] Input:
int((x*log(x^2)*log(2*log(x^2))^2+(20+4*x)*log(2*log(x^2))+((-x^2-6*x)*exp (x)+4*x^2+15*x)*log(x^2))/x/log(x^2),x)
Output:
- e**x*x - 5*e**x + log(2*log(x**2))**2*x + 5*log(2*log(x**2))**2 + 2*x** 2 + 15*x